Question

A sphere of radius R and charge 'Q' is placed inside an imaginary sphere of radius 2R such that the centres of two spheres coincide. The electric flux linked with the imaginary sphere is

• A. $\frac{4Q}{\varepsilon_0}$
• B. $\frac{2Q}{\varepsilon_0}$
• C. $\frac{Q}{\varepsilon_0}$
• D. $\frac{Q}{2\varepsilon_0}$

Hint:

To calculate the electric flux through a closed surface, apply Gauss's Law. The total electric flux depends only on the net charge enclosed within the Gaussian surface and is independent of the size or shape of the surface, as long as it encloses the charge.

Answer 1

The Correct Option is C

Step 1: Identify the Relevant Concept and Formula
The problem involves calculating the electric flux through a closed surface due to a charge enclosed within it. This is a direct application of Gauss's Law in electrostatics.
Gauss's Law states that the total electric flux ($\Phi_E$) through any closed surface (called a Gaussian surface) is equal to the total electric charge ($Q_{enclosed}$) enclosed within that surface divided by the permittivity of free space ($\varepsilon_0$).
The formula for electric flux according to Gauss's Law is: \[ \Phi_E = \frac{Q_{enclosed}}{\varepsilon_0} \] Step 2: Identify the Enclosed Charge and Gaussian Surface
We have a sphere of radius R carrying a charge 'Q'.
This sphere is placed inside an \textit{imaginary} sphere of radius 2R, and their centers coincide. The imaginary sphere acts as our Gaussian surface.
The charge 'Q' is entirely contained within the sphere of radius R, which in turn is entirely contained within the imaginary sphere of radius 2R.
Therefore, the total charge enclosed ($Q_{enclosed}$) by the imaginary sphere (Gaussian surface) is exactly the charge 'Q' present on the inner sphere. \[ Q_{enclosed} = Q \] The radius of the Gaussian surface (imaginary sphere, 2R) does not affect the total flux, as long as it encloses the charge. Step 3: Compute the Electric Flux
Substitute the identified enclosed charge into Gauss's Law formula: \[ \Phi_E = \frac{Q}{\varepsilon_0} \] Step 4: Verify the Result (Conceptual Understanding)
Gauss's Law is independent of the shape and size of the Gaussian surface, as long as it fully encloses the charge. The fact that the imaginary sphere has a radius of 2R (or any radius greater than R) does not change the total charge enclosed, which remains 'Q'. Therefore, the flux linked with the imaginary sphere will solely depend on the enclosed charge 'Q' and the permittivity of free space $\varepsilon_0$. Step 5: Analyze the Options
\begin{itemize} \item Option (1): $\frac{4Q}{\varepsilon_0}$. Incorrect. \item Option (2): $\frac{2Q}{\varepsilon_0}$. Incorrect. \item Option (3): $\frac{Q}{\varepsilon_0}$. Correct, as it matches our calculated electric flux based on Gauss's Law. \item Option (4): $\frac{Q}{2\varepsilon_0}$. Incorrect. \end{itemize}